CHAPTER 19 CURRENT, RESISTANCE AND ELECTROMOTIVE FORCEpeople.physics.tamu.edu/adair/phys202/CHAPTER...
Transcript of CHAPTER 19 CURRENT, RESISTANCE AND ELECTROMOTIVE FORCEpeople.physics.tamu.edu/adair/phys202/CHAPTER...
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CHAPTER 19
CURRENT, RESISTANCE AND
ELECTROMOTIVE FORCE
BASIC CONCEPTS
CURRENT and CURRENT DENSITY
RESISTANCE and RESISTIVITY
BATTERY INTERNAL RESISTANCE
ENERGY AND POWER IN CIRCUITS
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CURRENT IS THE MOVEMENT OF CHARGE
IN A MATERIAL
In some cases the objects that are moving
are positive.
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RESISTIVITY
Resistivity, ρ, changes with temperature.
���� = ���1 + �� − ����
Where �� is the value at 200 C.
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Some materials loose all resistivity –
Superconductors:
Discovered in 1911 by H. Kammerlingh
Onnes.
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RESISTANCE
Resistance is a characteristic of an object.
Resistance, R, is related to Resistivity, ρ, by
= ��
�
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The longer the object the higher the
resistance.
The larger the cross-section the smaller the
resistance.
If ρ is constant the total current through a
conductor is proportional to the voltage
across it.
� ∝ �
Or
� ∝ �
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The constant of proportionality is the
resistance, R.
� = �
Or
� = �
This is Ohm’s Law.
If a resistor obeys Ohm’s Law it is an Ohmic
resistor.
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CALCULATE RESISTANCE
A copper rod with cross section A has a
length L what is its resistance? Work for
� = 3.14�10� !� and � = 10!.
From Table 25.1 � = 1.72�10�$Ω!
Therefore
= ��
�= �1.72�10�$Ω!�
10!
3.14�10� !�
= 5.48�10��Ω
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The figure is of a 12 V battery with no
current in the circuit.
The voltmeter will read 12 V.
But if there is current:
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What is the current in the circuit?
� = �
� =�
=
12
4 + 2= 2�
What will the voltmeter read?
�()*+ = 12� − � ,-.)(-*/
�()*+ = 12� − �2���2Ω�
�()*+ = 12� − 4� = 8�
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DIRECT-CURRENT CIRCUITS
BASIC CONCEPTS
Resistors in circuits.
Kirchhoff’s Rules
Ammeters and Voltmeters
R-C Circuits
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In Chapter 24 we found equivalent
capacitance for combinations of capacitors.
1
0)1=
1
02+
1
0�456786987
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For resistors we need similar equations.
The current I through each R is same.
The voltage �*= = �*> + �>? + �?=
And � = �
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Therefore
� 2 + � � + � @ = � )1A,B*/)-.
Can replace the three resistors with one
resistor with resistance )1A,B*/)-.
)1A,B*/)-. = 2 + � + @456786987
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��R =��R
2Ω=12�
2Ω= 6�
Therefore there are 6A through the 1Ω
resistor.
�2T = �2R1Ω = 6�1Ω = 6�
The emf of the battery is
U = �2T + ��R = 6� + 12� = 18�
The current through the 6Ωresistor is
� R =U
6Ω=18�
6Ω= 3�
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Example:
Loop 1 (start at a)
−�� + U2 − �262 = 0
Loop 2 (start at c)
+U2 − �262 − ��6� − U� = 0
I1
I2
IR